Quick look over friction pulley system

  • Thread starter ngorecki
  • Start date
  • #1
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Homework Statement



Objects with masses m1 = 12.0 kg and m2 = 4.00 kg are connected by a light string that passes over a frictionless pulley. If, when the system starts from rest, m2 falls 1.00 m in 1.70 s, determine the coefficient of kinetic friction between m1 and the table.



Homework Equations



Fg = Fn = mg
F = ma
μ = Ff/Fn

The Attempt at a Solution



Fg = mg
Fg = 16*9.8
Fg = 156.8 = Fn

F = ma [m2]
= 4*9.8
= 39.2

ƩFx = m2 - Ff (sum of all forces in the x direction is = m2 - Frictional force)
= 39.2 - Ff
Ff = 39.2

μ = Ff/Fn
μ = 39.2/156.8
μ = .25

I tried entering the coefficient of kinetic friction into the "submit box" but the answer is incorrect.
could somebody please point out my mistake?
 

Answers and Replies

  • #2
ehild
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Is there a table in the problem??? A drawing would be nice, with free body diagram for both masses.

ehild
 
  • #3
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this is the picture i have
 

Attachments

  • p4_25.gif
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  • #4
ehild
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Nice picture:smile:.
Both blocks accelerate with the same amount as they are connected with a spring.
m2 falls 1 m in 1.7 s. What is its acceleration?

ehild
 
  • #5
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thanks :)
and acceleration is:
Δx = .5 a t^2
1 = .5(a)1.7^2
1 = 1.445a
.69 = a
 
  • #6
6
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Your mistake is in your calculation of normal force. Only m1 should factor into normal force. M2 experiences only gravity and tension :)
 
  • #7
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i multiply m1 by .69?
 
  • #8
ehild
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It is OK,
but do not round off during the calculations, keep one more digit as the number of the significant digits in the data. And show the unit. So a=0.6920 m/s2

The acceleration of m2 is know, so the net force on m2 is F=m2a. What forces act on m2?

ehild
 
  • #9
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gravity and tension?
 
  • #10
ehild
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i multiply m1 by .69?

If you multiply m1 with the common acceleration, you get the net horizontal force on m1. That force is the resultant of what forces?

Think of the string. Does it exert force?

ehild
 
  • #11
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the string which is connected to m2 pulls m1. and then there is friction on m1
 
  • #12
ehild
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gravity and tension?

Yes. Gravity and tension of the rope result in F=m2a. Can you find the magnitude of the tension?

ehild
 
  • #13
ehild
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the string which is connected to m2 pulls m1. and then there is friction on m1

Yes, what are the direction of these forces?

ehild
 
  • #14
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gravity on m2 = 39.2 N. the tension = 39.2 also?
and friction is in the "negative" direction
force from m2 is in the "positive" direction
 
  • #15
ehild
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gravity on m2 = 39.2 N. the tension = 39.2 also?
and friction is in the "negative" direction
force from m2 is in the "positive" direction

The tension in the rope acts upward on m2. See the picture, I added the forces. So the net force on m2 is m2g-T=m2a. What is T then?

ehild
 

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  • #16
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36.432 = t
 
  • #17
ehild
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36.432 = t

Yes, the tension is T=36.432 N.
Now what forces act on m1?

ehild
 
  • #18
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Tension and friction?
 
  • #19
ehild
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Yes. The tension is the same at both ends of the string. You know acceleration, mass and one force, tension. How much is the other force, the force of friction?

ehild
 
  • #20
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ƩFx = T - Ff
ƩFx = 36.432 - Ff
Ff = 36.432 N
 
  • #21
ehild
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Is ƩFx =0???? m1 is accelerating!


ehild
 
  • #22
37
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M1 = ma
M1 = 12 * .692
M1 = 8.304

M1 = T- Ff
8.304 = 36.432 - Ff
Ff + 8.034 = 36.432
Ff = 28.398

Is this accurate?
 
  • #23
ehild
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Why do you denote force by M?

So the net force on m1 is m1a=8.304 N, and it is equal to T-Ff. But what is 8.034?

ehild
 
  • #24
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the mass of m1 * acceleration = 8.304
then i take the mass of m1 * gravity = 117.6
u = Ff/Fn = 28.128/117.6 = .239

the answer is.239
 
  • #25
ehild
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Correct !

ehild
 

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